Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 21: Problem 67

The compound cisplatin, \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\), has been studied extensively as an antitumor agent. The reaction for the synthesis of cisplatin is: $$\mathrm{K}_{2} \mathrm{PtCl}_{4}(a q)+2 \mathrm{NH}_{3}(a q) \longrightarrow \mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}(s)+2 \mathrm{KCl}(a q)$$ Write the electron configuration for platinum ion in cisplatin. Most \(d^{8}\) transition metal ions exhibit square planar geometry. With this and the name in mind, draw the structure of cisplatin.

Short Answer

Expert verified
The electron configuration of the platinum ion in cisplatin is [Xe] 4f14 5d8. Cisplatin has a square planar geometry with the central Pt atom surrounded by two Cl- ions positioned together and two NH3 ligands positioned together on the other side of the square, approximately 90 degrees apart.

Step by step solution

01

Writing the electron configuration of the platinum ion in cisplatin

In the standard state, Platinum (Pt) has an electron configuration of [Xe] 4f14 5d9 6s1. In the reaction given, platinum in \(K_2 PtCl_4\) loses two electrons to form a Pt(II) ion. So the electron configuration of Pt(II) ends up as [Xe] 4f14 5d8 because two of the electrons in the 5d and 6s orbitals are removed forming a d8 configuration.
02

Drawing the structure of cisplatin

Because Pt(II) has a d8 configuration and most d8 transition metals have square planar geometry, cisplatin also exhibits similar geometry. This means that the central atom, Pt in this case, lies in the center with four ligands (NH3 and Cl in this case) attached at 90-degree angles from each other in a plane creating a square. The cis- in the name cisplatin represents that the same types of ligands (uptake Cl-) are beside each other. Therefore, the Cl- ions will be positioned next to each other(approximately 90 degrees apart) and the NH3 ligands will also be positioned together on the other side of the square.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Acetylacetone (see Exercise 69, part a), abbreviated acacH, is a bidentate ligand. It loses a proton and coordinates as acac \(^{-}\), as shown below: Acetylacetone reacts with an ethanol solution containing a salt of europium to give a compound that is \(40.1 \% \mathrm{C}\) and \(4.71 \% \mathrm{H}\) by mass. Combustion of \(0.286 \mathrm{~g}\) of the compound gives \(0.112 \mathrm{~g}\) \(\mathrm{Eu}_{2} \mathrm{O}_{3}\). Assuming the compound contains only \(\mathrm{C}, \mathrm{H}, \mathrm{O}\), and \(\mathrm{Eu}\), determine the formula of the compound formed from the reaction of acetylacetone and the europium salt. (Assume that the compound contains one europium ion.)

Draw all geometrical and linkage isomers of \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{NO}_{2}\right)_{2}\).

Which of the following ligands are capable of linkage isomerism? Explain your answer. $$\mathrm{SCN}^{-}, \mathrm{N}_{3}^{-}, \mathrm{NO}_{2}^{-}, \mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}, \mathrm{OCN}^{-}, \mathrm{I}^{-}$$

Hemoglobin (abbreviated \(\mathrm{Hb}\) ) is a protein that is responsible for the transport of oxygen in the blood of mammals. Each hemoglobin molecule contains four iron atoms that serve as the binding sites for \(\mathrm{O}_{2}\) molecules. The oxygen binding is \(\mathrm{pH}\) dependent. The relevant equilibrium reaction is $$\mathrm{HbH}_{4}{ }^{4+}(a q)+4 \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)+4 \mathrm{H}^{+}(a q)$$ Use Le Châtelier's principle to answer the following. a. What form of hemoglobin, \(\mathrm{HbH}_{4}^{4+}\) or \(\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}\), is favored in the lungs? What form is favored in the cells? b. When a person hyperventilates, the concentration of \(\mathrm{CO}_{2}\) in the blood decreases. How does this affect the oxygenbinding equilibrium? How does breathing into a paper bag help to counteract this effect? (Hint: \(\mathrm{CO}_{2}\) reacts with water to produce carbonic acid.) c. When a person has suffered a cardiac arrest, an injection of a sodium bicarbonate solution is given. Why is this step necessary?

Henry Taube, 1983 Nobel Prize winner in chemistry, has studied the mechanisms of the oxidation-reduction reactions of transition metal complexes. In one experiment he and his students studied the following reaction: \(\begin{aligned} \mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}(a q)+& \mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}^{2+}(a q) \\\ \longrightarrow & \mathrm{Cr}(\mathrm{III}) \text { complexes }+\mathrm{Co}(\mathrm{II}) \text { complexes } \end{aligned}\) Chromium(III) and cobalt(III) complexes are substitutionally inert (no exchange of ligands) under conditions of the experiment. Chromium(II) and cobalt(II) complexes can exchange ligands very rapidly. One of the products of the reaction is \(\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cl}^{2+} .\) Is this consistent with the reaction proceeding through formation of \(\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cr}-\mathrm{Cl}-\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\) as an intermediate? Explain.
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Analysis

Read Explanation

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free